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22-Agric-B8 Food Process Engineering (Part 1) · May 2014

Question 9 of 10: Double-Effect Evaporator, Forward Feed — Solids Leaving the First Effect

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-B8 Food Process Engineering (Part 1), National Exams May 2014 — a three-hour open-book exam (any non-communicating calculator permitted). Ten questions are set in four sections (I–IV), each with a "do one/any N of M" instruction; a candidate following the choice rules answers six questions for a 100-mark paper. All ten are worked here so the set is a complete study resource.

Reference texts. R.T. Toledo, Fundamentals of Food Process Engineering, 3rd ed. (thermal-process lethality, D and z values, Ball/Stumbo process calculation, aseptic holding-tube residence time, evaporator design — this is the exam's own appendix source); C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed. (evaporator heat and mass balances, multiple-effect steam economy); R.P. Singh and D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation, modified Plank and Cleland-Earle equations, unsteady-state heat transfer in canned foods); A.C. Cleland, Food Refrigeration Processes: Analysis, Design and Simulation (Plank/Cleland-Earle freezing-time correlations); F.P. Incropera and D.P. DeWitt, Fundamentals of Heat and Mass Transfer (transient conduction, Heisler charts, composite-wall resistance).

Check: this paper's four roman-numeral section headers ("I. Heat transfer", "II. Food freezing and freeze concentration", "III. Thermal processing", "IV. Several assumptions (retort come-up correction factor, reference temperature for spore D-values, evaporator steam temperature reused for Question 9) are flagged inline where the source leaves a value implicit.

Question 9: Double-Effect Evaporator, Forward Feed — Solids Leaving the First Effect (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Forward-feed double-effect evaporator data
QuantitySymbolValue
Feed basis\(F\)100 kg (10 kg solids + 90 kg water)
Feed solids, temperature\(x_F,\,T_F\)10%, 55°C
Final product solids, temperature\(x_{mc},\,T_{mc}\)30%, 70°C
Effect 1 / Effect 2 temperature—75°C / 70°C
Specific heat, solids / water\(C_{p,s},\,C_{p,w}\)2.09, 4.186 kJ/(kg·K)

Find. The solids content of the liquid stream A leaving the first effect.

Effect 1(75 C)Effect 2(70 C)Feed F=100 kg10% solids, 55 CSteam S1=37.61 kg, 100 CA(liquid)=66.86 kg14.96% solids, 75 CV1=33.14 kg vapour, 75 C -> to Effect-2 coil<- V1 vapour from Effect-1 coilmc=33.33 kg30% solids, 70 CB(V2)=33.52 kgvapour, 70 C
Fig. E — forward-feed double-effect evaporator: both the liquid (feed→A→mc) and the heating vapour (Effect 1→Effect 2 coil) move in the SAME direction, hot effect to cold effect.

Approach. As in Question 8, an overall solids balance fixes the final product mass. Because the stream leaving Effect 1 arrives at Effect 2 already at 75°C (hotter than Effect 2's own 70°C boiling point), it partly flash-evaporates on its own sensible-heat drop in addition to whatever the coil supplies — both terms enter the Effect 2 energy balance. The vapour \(A\) evaporated in Effect 1 is the only unknown once the mass and energy balances are combined, so it is solved by root-finding (the intermediate stream's own specific heat depends on \(A\), making the equation nonlinear).

  1. Overall solids balance. \(m_c = 10/0.30=\boxed{33.33\ \text{kg}}\); feed specific heat \(C_{pF}=0.10(2.09)+0.90(4.186)=\boxed{3.977\ \text{kJ/(kg}\cdot\text{K)}}\).
  2. Mass balance, stream A and Effect 2 vapour B. Stream A leaving Effect 1 has mass \((100-A)\) and carries all 10 kg of solids, so its own specific heat is \(C_{pA}=\dfrac{10}{100-A}(2.09)+\left(1-\dfrac{10}{100-A}\right)(4.186)\). Overall, \(B=(100-A)-m_c=66.67-A\).
  3. Energy balance on Effect 2 (feed already hot — flash + coil heat). Vapour from Effect 1 (75°C) condenses in Effect 2's coil, AND stream A itself gives up sensible heat cooling from 75 to 70°C as it flashes into the lower-pressure effect: \(A\,h_{fg,75} + (100-A)\,C_{pA}(75-70) = B\,h_{fg,70}\), with \(h_{fg,75}=2321.4\), \(h_{fg,70}=2333.8\ \text{kJ/kg}\).
  4. Solve for \(A\) (root-finding, since \(C_{pA}\) itself depends on \(A\)). Substituting \(B=66.67-A\) and solving numerically: \(A=\boxed{33.14\ \text{kg}}\), so the stream leaving Effect 1 has mass \(100-A=\boxed{66.86\ \text{kg}}\) and \(B=66.67-33.14=\boxed{33.52\ \text{kg}}\).
  5. Solids content leaving Effect 1. \(x_A = \dfrac{10}{66.86}=\boxed{14.96\%}\) — again between the 10% feed and 30% final concentrations, and close to (but not identical to) Question 8's reverse-feed intermediate of 14.70%, since the two configurations expose the intermediate stream to different heat loads.
Final results
QuantityValue
Vapour evaporated in Effect 1, \(A\)33.14 kg
Mass of stream leaving Effect 166.86 kg
Solids content leaving Effect 1, \(x_A\)14.96%
Vapour \(B\) from Effect 233.52 kg
Final concentrate \(m_c\)33.33 kg, 30% solids
Check: two assumptions fill gaps the extracted figure leaves implicit: (1) the feed basis is 100 kg, read from the figure's own "10 kg solid + (90−A) kg water" label, consistent with 10% solids of a 100 kg feed; (2) the live steam temperature is taken as 100°C, reusing Question 8's value for the same evaporator system, since no steam temperature is printed on this figure.